Box on Box on Box free body diagram

wil3

Okay, so say that I stack three boxes, one on top of another. There is friction between all surface, including the bottom box and the ground.

If I apply a force to the center box only, is there any set of circumstances under which the top box will slip? I know that if the applied force is large enough, static friction between the bottom and center box will be overcome and the middle box will slip over the bottom box. Is there a way for the top box to slip out as well? I am trying to draw a free body diagram, but the only x-axis force I find on the top box is friction, which always points in the direction of the force applied to the center box. I know that the applied force can be great enough to overcome the static friction between the top and middle boxes, but it seems to me like the top box will still accelerate some in hte direction of the applied force. Is this accurate?

Think about this in terms of a playing card deck- if I take a deck of cards and set it on a table, and I pull out on random card halfway, if I flick it hard enough it will go flying out of hte deck without disturbing the cards above or below it.

jack action

Yes.

Middle box:

[tex]F_{pull} - F_{f \ bottom} - F_{f \ top} = m_{middle} a_{middle}[/tex]

Top box:

[tex]F_{f \ top} = m_{top} a_{top}[/tex]

Both accelerations are in the same direction but they are not the same magnitude. And if the middle one accelerate fast enough wrt the top box, the top box will seem immobile (it won't have time to travel very far before the middle box is gone). Similarly, for the bottom box:

[tex]F_{f \ bottom} + F_{f \ ground} = m_{bottom} a_{bottom}[/tex]

wil3

Thank you very much! Of all of the questions I have ever posted on Physics Forums, this is one of the few that I have received a complete, well-worded, and properly explained answer to.

jack action

Thank you. I answer as much for myself than for others. It's exercise for the brain.